1eChemical formulae, equations and calculations

A balanced equation says that atoms are never created or destroyed. Whatever goes in must come out, atom for atom.

The one rule

You may change the big number in front of a formula. You may never change the formula itself.

Water is H₂O. Writing H₂O₂ to make the oxygen balance does not fix the equation — it replaces water with hydrogen peroxide, a different substance. If an equation will not balance, the coefficients are wrong, not the formulae.

A method that works every time

Take methane burning: CH₄ + O₂ → CO₂ + H₂O

1. Carbon first. One on each side. Done.
2. Hydrogen next. Four on the left, two on the right. Put a 2 in front of H₂O.
3. Oxygen last. Now the right has 2 (in CO₂) + 2 (in 2H₂O) = 4. So the left needs 2O₂.

CH₄ + 2O₂ → CO₂ + 2H₂O

Balance oxygen last, because it usually appears in more than one product. Doing it first means redoing it.

State symbols

The distinction that costs marks is (l) versus (aq). Pure water is (l). Salt dissolved in water is (aq), not (l) — the (aq) is what tells you it is a solution.

Word equations

If a question asks for a word equation, give words. Examiners report candidates offering a balanced symbol equation instead, which does not answer the question asked and earns nothing — even when it is correct.

Relative formula mass, Mr

Add up the relative atomic mass of every atom in the formula.

For CO₂: 12 + 16 + 16 = 44. The subscript 2 applies only to the oxygen.

Brackets

A number after a bracket multiplies everything inside it.

For Mg(NO₃)₂: the bracket holds 14 + (3 × 16) = 62. Doubled, that is 124. Add magnesium: 24 + 124 = 148.

This is one of the most frequently penalised slips in the whole calculation topic. Read Pb(NO₃)₂ as a shopping list: one lead ion, and two whole nitrate groups — each of which carries its own nitrogen and three oxygens. Count the atoms out loud if it helps: 1 Pb, 2 N, 6 O.

The mole

The mole is the unit for amount of substance. It counts particles, in the way a dozen counts objects.

One mole of any substance has a mass in grams equal to its Mr. So one mole of water weighs 18 g; one mole of calcium carbonate weighs 100 g.

Converting between mass and moles

amount (mol) = mass (g) ÷ Mr

mass (g) = amount (mol) × Mr

Work out the Mr first and write it down. If the Mr is wrong every later step fails, but a correct Mr still earns its own mark even when the arithmetic afterwards goes astray.

Amount is not mass

One mole of carbon and one mole of magnesium contain the same number of atoms but have different masses — 12 g and 24 g — because a magnesium atom is twice as heavy.

Equal amounts do not mean equal masses. Keeping amount and mass separate in your head is what makes the rest of this sub-topic work.

The three-step route

Every reacting-mass calculation follows the same path:

1. Mass → moles for the substance you know. Divide by its Mr.
2. Moles → moles using the ratio in the balanced equation.
3. Moles → mass for the substance you want. Multiply by its Mr.

A worked example

CaCO₃ → CaO + CO₂. What mass of CaO comes from 50 g of CaCO₃?

1. Mr of CaCO₃ = 100, so amount = 50 ÷ 100 = 0.5 mol
2. The ratio is 1:1, so 0.5 mol of CaO forms
3. Mr of CaO = 56, so mass = 0.5 × 56 = 28 g

Use step 2 even when the ratio is 1:1. It costs nothing there and it is the step that saves you when the ratio is 1:2.

Watch the coefficients

For Fe₂O₃ + 3CO → 2Fe + 3CO₂, one mole of iron(III) oxide gives two moles of iron. Missing that 2 halves the answer.

Examiners also report candidates using the wrong Mr entirely — for instance treating iron as 112 (which is 2 × 56) and then dividing by 2. Write the Mr of each substance out before you start.

"In excess"

When a question says one reactant is in excess, base the calculation on the other one. The excess reactant is not the limiting factor, so its amount tells you nothing about how much product forms.

Percentage yield

percentage yield = (actual yield ÷ theoretical yield) × 100

The theoretical yield is what the equation predicts. The actual yield is what was collected.

Two checks:

• The actual yield goes on top. A result above 100% means the fraction is inverted.
• Multiply by 100 at the end. Examiners report candidates completing the division correctly and then losing the final mark by not converting it to a percentage.

Yields fall short for specific, nameable reasons: product lost when transferring between containers, product left on the filter paper, or a reaction that did not go to completion. "Human error" is not creditworthy.

Two kinds of formula

Ethene is C₂H₄ molecularly but CH₂ empirically, because 2:4 simplifies to 1:2. Water is H₂O both ways, since 2:1 is already simplest.

Divide by the highest common factor. For C₄H₁₀ that factor is 2, giving C₂H₅ — not 4, because 10 does not divide by 4.

Finding an empirical formula from data

1. Convert each mass to moles (mass ÷ Ar).
2. Divide every result by the smallest of them.
3. Round to whole numbers.

Percentages can be used directly as masses — treat them as the masses in 100 g of the compound. There is no conversion step.

From empirical to molecular

Divide the relative molecular mass by the empirical formula mass, then multiply every subscript by that number.

CH₂ has a formula mass of 14. If Mr is 56, then 56 ÷ 14 = 4, and the molecular formula is C₄H₈. Multiply both subscripts — multiplying only the carbon is a common slip.

Water of crystallisation

Hydrated salts hold water inside their crystals. Heating drives it off, and the loss in mass is the mass of that water.

For MgSO₄·xH₂O, find x like this:

1. Mass of water = mass before − mass after
2. Moles of water = that mass ÷ 18
3. Moles of anhydrous salt = its mass ÷ its Mr
4. x = moles of water ÷ moles of salt

x comes out a whole number. If it does not, check step 1 — the subtraction is usually where it went wrong.

Finding a formula by combustion

Burning magnesium in a crucible gives the formula of magnesium oxide by weighing before and after.

Two practical points carry marks:

• Lift the lid at intervals to let oxygen in, so all the magnesium reacts — but do not leave it off, or white magnesium oxide smoke escapes and the product mass comes out too low.
• Heat to constant mass. Repeating the heat-cool-weigh cycle until the mass stops changing is the only evidence that the reaction has finished.

Separate Chemistry only.

Solutions

amount (mol) = concentration (mol/dm³) × volume (dm³)

The trap is units. Concentration is per dm³, and volumes in questions are usually given in cm³.

cm³ ÷ 1000 = dm³

So 250 cm³ is 0.250 dm³. Forgetting that division makes the answer a thousand times too large, which is the commonest error in the whole topic.

Worked example

4.0 g of NaOH dissolved and made up to 500 cm³. What is the concentration?

1. Mr of NaOH = 40, so amount = 4.0 ÷ 40 = 0.10 mol
2. Volume = 500 ÷ 1000 = 0.500 dm³
3. Concentration = 0.10 ÷ 0.500 = 0.20 mol/dm³

Two conversions before the final division — grams to moles, and cm³ to dm³. Missing either is where marks go.

Gases

At room temperature and pressure, one mole of any gas occupies 24 dm³, or 24 000 cm³.

amount (mol) = volume ÷ molar volume

The molar volume is the same for every gas. Hydrogen and carbon dioxide occupy identical volumes per mole, because in a gas the particles are so far apart that volume depends on how many there are, not on their size or mass.

Match your units

Use 24 with dm³ and 24 000 with cm³. Mixing them is a factor-of-a-thousand error.

Where the molar volume applies

Only to the gas. In a calculation like magnesium reacting with acid to give hydrogen, the route is:

1. Mass of magnesium → moles
2. Equation ratio → moles of hydrogen
3. Moles of hydrogen × 24 → volume in dm³

Applying 24 to the magnesium, which is a solid, is a costly and frequent mistake.

The formula is found by measuring the mass of metal and the mass of oxygen combined with it. There are two routes: add oxygen to a metal, or take oxygen away from an oxide.

Route 1 — burning magnesium

Apparatus

Crucible with lid, pipeclay triangle, tripod, Bunsen, tongs, balance, emery paper.

Method

1. Clean a coil of magnesium ribbon with emery paper.
2. Weigh the empty crucible and lid. Add the magnesium and reweigh.
3. Heat strongly with the lid on.
4. Lift the lid briefly and often, replacing it each time.
5. Heat to constant mass, then cool and reweigh.

The calculation

Divide each mass by its relative atomic mass, then divide both answers by the smaller. A ratio near 1 : 1 gives MgO.

The mass of oxygen is the increase in mass, never the final mass.

Route 2 — reducing copper(II) oxide

Copper(II) oxide in a reduction tube is heated while a stream of hydrogen (or natural gas) passes over it. The black oxide turns pink-brown as copper forms, and the mass falls.

The safety sequence is the method here, and the order is not optional:

1. Pass the gas through first, long enough to flush all the air out of the tube.
2. Only then light the burner.
3. Keep heating until cool, so gas is still flowing while the hot copper is exposed.

Hydrogen mixed with air explodes when ignited. Flushing first removes the air; keeping the gas flowing while cooling stops air being drawn back over hot copper, which would reoxidise it and ruin the result as well.

The excess gas leaving the tube is burnt off at a jet, not released into the room.

Here the mass of oxygen is the decrease. The reasoning is identical, run in reverse.

Why each step

Clean the ribbon. Magnesium tarnishes in air, so an uncleaned coil already carries oxygen. That oxygen was never weighed as part of the gain, and the ratio comes out wrong.

Lift the lid. Air must reach the metal or the reaction cannot finish — but the white magnesium oxide leaves as smoke if the crucible stays open, and any that escapes is missing from the final mass.

Constant mass. Proof the reaction is complete. Stopping early leaves unreacted magnesium and understates the oxygen.

Why results come out imperfect

A real experiment rarely gives exactly 1 : 1, and being able to say which way the error pushes the answer is what the question is testing.

Where marks are lost

• Using the final mass as the mass of oxygen.
• Dividing by 24 for oxygen or 16 for magnesium — the relative atomic masses swapped.
• Rounding the ratio to whole numbers before checking it is genuinely close to one.